Problem tracing -Code Output

[NovaPrimeveera]

Dear C++ Community

I do not seem to understand, the following code`s output even after tracing it.

  void f(int* &c, int* d){
     *c =16; 
     c = c+2;
     *d = (*c)%3;
     d = c;
     *d = 2*(*(d-1));
 }

 int main(){
    int a[5] = {5, 10, 15, 20, 25};
    int* p = a;
    int* q = &a[4];
    f(p,q);
    for (int i=0; i<5; i++)
       cout << a[i] << " ";
       cout << "\n" << *p;
       cout << "\n" << *q;
    return 0;
}

Code output:
============

16 10 20 20 0
20
0

-----------------------------------------------------------------------------
My confusion:
=============

When tracing the code I got 16, but my answer I kept leaving a line due to "\n". I am not sure why the answer only leaves a line after printing five values (as the last two cout, seem to be included in the for loop ).

My answer:
==========

16
16
1
10
16
1
15
16
1
20
16
1
1
16
1

[zapshe]

Line 17 will give an error, I assume you copied the code by hand you meant "<<" not ",,"

Either way, the reason it'll print out 5 numbers before starting a new line is because your for-loop is only looping through line 15 I suspect. Check on your code that the bracket on line 14 actually exists and ends somewhere.

[lastchance]

@NovaPrimeVeera, did you mean this?

#include <iostream>
using namespace std;
 
void f(int* &c, int* d){  // on entry, c points to the [0] (i.e. first) and d to the [4] (i.e. last) element
    *c =16;               // array becomes { 16, 10, 15, 20, 25 }
    c = c+2;              // c now points to the [2] element (i.e. the third)    
    *d = (*c)%3;          // the last element becomes 15 % 3, which is 0; array now { 16, 10, 15, 20, 0 }
    d = c;                // both pointers are now pointing to the middle element
    *d = 2*(*(d-1));      // the middle element becomes twice the [1] (i.e. second ) element; array now { 16, 10, 20, 20, 0 }    
}

int main(){
   int a[5] = {5, 10, 15, 20, 25};
   int* p = a;
   int* q = &a[4];
   f(p,q);                // after this, array is { 16, 10, 20, 20, 0 };
                          // p passed by reference: points to [2] element on return; q passed by value: still points to last element
   for (int i=0; i<5; i++)
      cout << a[i] << " ";      // The loop ends HERE
       
   cout << "\n" << *p;    // the [2] (i.e. third) element is 20
   cout << "\n" << *q;    // the last element is 0
   return 0;
}

16 10 20 20 0 
20
0

Other than as a work-up to assembling flat-pack furniture, is there any particular purpose to this code?

[NovaPrimeveera]

Nope, our lecturer designed that code for us to trace it, but I did not understand his answer for the tracing. Thank you for the clarification.

[salem c]

Rather than get lost in alphabet soup, rename variables to be something meaningful (or at least as meaningful as you understand). If you get a better idea, then rename them again.

#include <iostream>
using namespace std;
 
void f(int* &startOfArrayRef, int* endOfArray){ // on entry, startOfArrayRef points to the [0] (i.e. first) and endOfArray to the [4] (i.e. last) element
    *startOfArrayRef = 16;                      // array becomes { 16, 10, 15, 20, 25 }
    startOfArrayRef = startOfArrayRef+2;        // startOfArrayRef now points to the [2] element (i.e. the third)    
    *endOfArray = (*startOfArrayRef)%3;         // the last element becomes 15 % 3, which is 0; array now { 16, 10, 15, 20, 0 }
    endOfArray = startOfArrayRef;               // both pointers are now pointing to the middle element
    *endOfArray = 2*(*(endOfArray-1));          // the middle element becomes twice the [1] (i.e. second ) element; array now { 16, 10, 20, 20, 0 }    
}

int main(){
   int a[5] = {5, 10, 15, 20, 25};
   int* startOfArray = a;
   int* endOfArray = &a[4];
   f(startOfArray,endOfArray);                  // after this, array is { 16, 10, 20, 20, 0 };
                                                // startOfArray passed by reference: points to [2] element on return; 
                                                // endOfArray passed by value: still points to last element
   for (int i=0; i<5; i++)
      cout << a[i] << " ";      // The loop ends HERE
       
   cout << "\n" << *startOfArray;               // the [2] (i.e. third) element is 20
   cout << "\n" << *endOfArray;                 // the last element is 0
   return 0;
}